## Strength of Materials |

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Page 105

As we saw in the preceding article , the effect of the loads to the left of this

segment

this segment produce the slightly different values of shear and moment V + DV

and M + ...

As we saw in the preceding article , the effect of the loads to the left of this

segment

**reduces**to the shear V and the moment M , and the loads to the right ofthis segment produce the slightly different values of shear and moment V + DV

and M + ...

Page 113

The concentrated load of 120 lb at E

we locate the positions of zero shear at F and G on the shear diagram , consider

the effect of narrowing the distance over which the reaction R2 is distributed .

The concentrated load of 120 lb at E

**reduces**the shear abruptly to zero . Beforewe locate the positions of zero shear at F and G on the shear diagram , consider

the effect of narrowing the distance over which the reaction R2 is distributed .

Page 559

... ( b ) by its value from Eq . ( a ) , we obtain Iu = S ( y2 cos ” a – 2xy sin a cos a +

x ? sino a ) dA Since 1 , = Syề da , 1 , = Sx ? da , and Pzy = Sxy dA , this

to - In = I x cosa a + I , sin ? a – Prysin 2a If the relations 1 + cos 2a and , sino a =

?

... ( b ) by its value from Eq . ( a ) , we obtain Iu = S ( y2 cos ” a – 2xy sin a cos a +

x ? sino a ) dA Since 1 , = Syề da , 1 , = Sx ? da , and Pzy = Sxy dA , this

**reduces**to - In = I x cosa a + I , sin ? a – Prysin 2a If the relations 1 + cos 2a and , sino a =

?

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### Contents

List of Symbols and Abbreviations | xvi |

SIMPLE STRAIN | 26 |

TORSION | 60 |

Copyright | |

18 other sections not shown

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### Common terms and phrases

acting actual allowable angle applied assumed axes axial axis beam beam shown bending bending moment cantilever carries caused centroid circle column compressive compressive stress compute concentrated concrete consider constant couple cross section deflection deformation Determine developed diameter direction distance distributed load effect elastic curve element equal equation equivalent expressed flange flexural stress force formula ft-lb given gives Hence horizontal ILLUSTRATIVE inertia joint lb/ft length limit load material maximum maximum shearing method midspan moments negative neutral axis normal obtain occurs plane plate positive principal Prob PROBLEMS produce radius reaction reduces reference reinforced relation resisting respect resultant rivet segment shaft shearing stress shown in Fig shows slope Solution Solve span steel strain strength supported Table tangent tensile thickness torsional uniformly varies vertical weight weld yield zero